CMPT 225 - Simon Fraser University · 2017. 10. 15. · Choosing an appropriate algorithm can make...

CMPT 225

Measuring algorithm efficiency▪ Timing

▪ Counting

Cost functions▪ Cases

▪ Best case

▪ Average case

▪ Worst case

▪ Searching

▪ Sorting

O Notation▪ O notation's mathematical basis

▪ O notation classes

▪ and notations

John Edgar 2

Algorithms can be described in terms of time and space efficiency

Time

▪ How long, in ms, does my algorithm take to solve a particular problem?

▪ How much does the time increase as the problem size increases?

▪ How does my algorithm compare to other algorithms?

Space

▪ How much memory space does my algorithm require to solve the problem?

John Edgar 4

Choosing an appropriate algorithm can make a

significant difference in the usability of a system

▪ Government and corporate databases with many millions

of records, which are accessed frequently

▪ Online search engines and media platforms

▪ Big data

▪ Real time systems where near instantaneous response is

required

▪ From air traffic control systems to computer games

John Edgar 5

There are often many ways to solve a problem

▪ Different algorithms that produce the same results▪ e.g. there are numerous sorting algorithms

We are usually interested in how an algorithm performs when its input is large

▪ In practice, with today's hardware, most algorithms will perform well with small input

▪ There are exceptions to this, such as the Traveling Salesman Problem▪ Or the recursive Fibonacci algorithm presented previously …

John Edgar 6

It is possible to count the number of operations that an algorithm performs

▪ By a careful visual walkthrough of the algorithm or by

▪ Inserting code in the algorithm to count and print the number of times that each line executes

It is also possible to time algorithms

▪ Compare system time before and after running an algorithm▪ More sophisticated timer classes exist

▪ Simply timing an algorithm may ignore a variety of issues

John Edgar 7

profiling

It may be useful to time how long an algorithm takes to rum

▪ In some cases it may be essential to know how long an algorithm takes on a particular system

▪ e.g. air traffic control systems

▪ Running time may be a strict requirement for an application

But is this a good general comparison method?

▪ Running time is affected by a number of factors other than algorithm efficiency

John Edgar 8

CPU speed Amount of main memory Specialized hardware (e.g. graphics card) Operating system System configuration (e.g. virtual memory) Programming language Algorithm implementation Other programs System tasks (e.g. memory management) …

John Edgar 9

Instead of timing an algorithm, count the number of instructions that it performs

The number of instructions performed may vary based on

▪ The size of the input

▪ The organization of the input

The number of instructions can be written as a cost function on the input size

John Edgar 10

John Edgar 11

Operations performed on an array of length 10

|

declare and initialize i

perform comparison, print array element, and

increment i:10 times

||| ||| ||| ||| ||| ||| ||| ||| ||| ||| |

make comparison when i = 10

32 operations

void printArray(int arr[], int n){

for (int i = 0; i < n; ++i){

cout << arr[i] << endl;

}}

Instead of choosing a particular input size we will express a cost function for input of size n

▪ We assume that the running time, t, of an algorithm is proportional to the number of operations

Express t as a function of n

▪ Where t is the time required to process the data using some algorithm A

▪ Denote a cost function as tA(n)

▪ i.e. the running time of algorithm A, with input size n

John Edgar 12

John Edgar 13

Operations performed on an array of length n

1

declare and initialize i

perform comparison, print array element, and

increment i: n times

3n 1

make comparison when i = n

t = 3n + 2

void printArray(int arr[], int n){

for (int i = 0; i < n; ++i){

cout << arr[i] << endl;

}}

In the example we assumed two things

▪ Neither of which are strictly true …

Any C++ statement counts as a single operation

▪ Unless it is a function call

That all operations take the same amount of time

▪ Some fundamental operations are faster than others

▪ What is a fundamental operation in a high level language is multiple operations in assembly

These are both simplifying assumptions

John Edgar 14

The number of operations often varies based on the size of the input

▪ Though not always – consider array lookup

In addition algorithm performance may vary based on the organization of the input

▪ For example consider searching a large array

▪ If the target is the first item in the array the search will be very fast

John Edgar 15

Algorithm efficiency is often calculated for three broad cases of input

▪ Best case

▪ Average (or “usual”) case

▪ Worst case

This analysis considers how performance varies for different inputs of the same size

John Edgar 16

It can be difficult to determine the exact number of operations performed by an algorithm▪ Though it is often still useful to do so

An alternative to counting all instructions is to focus on an algorithm's barometer instruction▪ The barometer instruction is the instruction that is executed

the most number of times in an algorithm

▪ The number of times that the barometer instruction is executed is usually proportional to its running time

John Edgar 17

It is often useful to find out whether or not a list contains a particular item

▪ Such a search can either return true or false

▪ Or the position of the item in the list

If the array isn't sorted use linear search

▪ Start with the first item, and go through the array comparing each item to the target

▪ If the target item is found return true (or the index of the target element)

John Edgar 19

int linearSearch(int arr[], int n, int x){for (int i=0; i < n; i++){

if(arr[i] == x){return i;

} } //forreturn -1; //target not found

}

John Edgar 20

The function returns as soon as the target item is found

return -1 to indicate that the item has not been found

tlinear search = 3n+2Worst case cost function:

Search an array of n items The barometer instruction is equality checking (or

comparisons for short)

▪ arr[i] == x;

▪ There are actually two other barometer instructions▪ What are they?

How many comparisons does linear search perform?

John Edgar 21

int linearSearch(int arr[], int n, int x){for (int i=0; i < n; i++){

if(arr[i] == x){return i;

} } //forreturn -1; //target not found

}

look for barometer operations inloops, or functions, or recursive calls

the barometer operation is the mostfrequently executed operation

Best case

▪ The target is the first element of the array

▪ Makes 1 comparison

Worst case

▪ The target is not in the array or

▪ The target is at the last position in the array

▪ Makes n comparisons in either case

Average case

▪ Is it (best case + worst case) / 2, i.e. (n + 1) / 2?

John Edgar 22

There are two situations when the worst case occurs

▪ When the target is the last item in the array

▪ When the target is not there at all

To calculate the average cost we need to know how

often these two situations arise

▪ We can make assumptions about this

▪ Though these assumptions may not hold for a particular

use of linear search

John Edgar 23

A1: The target is not in the array half the time

▪ Therefore half the time the entire array has to be checked to determine this

A2: There is an equal probability of the target being at any array location

▪ If it is in the array

▪ That is, there is a probability of 1/n that the target is at some location i

John Edgar 24

Work done if the target is not in the array

▪ n comparisons

▪ This occurs with probability of 0.5 (A1)

John Edgar 25

Work done if target is in the array:

▪ 1 comparison if target is at the 1st location▪ Occurs with probability 1/n (A2)

▪ 2 comparisons if target is at the 2nd location▪ Also occurs with probability 1/n

▪ i comparisons if target is at the ith location

Take the weighted average of the values to find the total expected number of comparisons (E)

▪ E = 1*1/n + 2*1/n + 3*1/n + … + n * 1/n or

▪ E = (n + 1) / 2

John Edgar 26

Target is not in the array: n comparisons Target is in the array (n + 1) / 2 comparisons Take a weighted average of the two amounts:

▪ = (n * ½) + ((n + 1) / 2 * ½)

▪ = (n / 2) + ((n + 1) / 4)

▪ = (2n / 4) + ((n + 1) / 4)

▪ = (3n + 1) / 4

Therefore, on average, we expect linear search to perform (3n + 1) / 4 comparisons

John Edgar 27

If we sort the target array first we can change the linear search average cost to approximately n / 2

▪ Once a value equal to or greater than the target is found the search can end

▪ So, if a sequence contains 8 items, on average, linear search compares 4 of them,

▪ If a sequence contains 1,000,000 items, linear search compares 500,000 of them, etc.

However, if the array is sorted, it is possible to do much better than this by using binary search

John Edgar 28

int binarySearch(int arr[], int n, int x){int low = 0;int high = n - 1;int mid = 0;while (low <= high){

mid = (low + high) / 2;if(x == arr[mid]){

return mid; } else if(x > arr[mid]){

low = mid + 1;} else { //x < arr[mid]

high = mid - 1;}

} //whilereturn -1; //target not found

}John Edgar 29

Index of the last element in the array

Note: if, else if, else

The algorithm consists of three parts

▪ Initialization (setting lower and upper)

▪ While loop including a return statement on success

▪ Return statement which executes on failure

Initialization and return on failure require the same amount of work regardless of input size

The number of times that the while loop iterates depends on the size of the input

John Edgar 30

The while loop contains an if, else if, else statement The first if condition is met when the target is found

▪ And is therefore performed at most once each time the algorithm is run

The algorithm usually performs 5 operations for each iteration of the while loop

▪ Checking the while condition

▪ Assignment to mid

▪ Equality comparison with target

▪ Inequality comparison

▪ One other operation (setting either lower or upper)

John Edgar 31

Barometer instructions

In the best case the target is the midpoint element of the array

▪ Requiring just one iteration of the while loop

John Edgar 32

index 0 1 2 3 4 5 6 7

arr 1 3 7 11 13 17 19 23

mid = (0 + 7) / 2 = 3

binary search (arr, 11)

What is the worst case for binary search?

▪ Either the target is not in the array, or

▪ It is found when the search space consists of one element

How many times does the while loop iterate in the worst case?

John Edgar 33

index 0 1 2 3 4 5 6 7

arr 1 3 7 11 13 17 19 23

mid =

binary search (arr, 20)

(0 + 7) / 2 = 3 (4 + 7) / 2 = 5 (6 + 7) / 2 = 6 (7 + 7) / 2 = 7

Each iteration of the while loop halves the search space▪ For simplicity assume that n is a power of 2

▪ So n = 2k (e.g. if n = 128, k = 7 or if n = 8, k = 3)

How large is the search space?▪ After the first iteration the search space is n/2

▪ After the second iteration the search space is n/4

▪ After the kth iteration the search space consists of just one element▪ Note that as n = 2k, k = log2n

▪ The search space of size 1 still needs to be checked

▪ Therefore at most log2n + 1 iterations of the while loop are made in the worst case

John Edgar 34

n/2k = n/n = 1

or n/21

or n/22

tbinary search = 5(log2(n)+1)+4Cost function:

Is the average case more like the best case or the worst case?

▪ What is the chance that an array element is the target

▪ 1/n the first time through the loop

▪ 1/(n/2) the second time through the loop

▪ … and so on …

It is more likely that the target will be found as the search space becomes small

▪ That is, when the while loop nears its final iteration

▪ We can conclude that the average case is more like the worst case than the best case

John Edgar 35

John Edgar 36

n Linear Search

(3n+1)/4

Binary Search

log2(n)+1

10 8 4

100 76 8

1,000 751 11

10,000 7,501 14

100,000 75,001 18

1,000,000 750,001 21

10,000,000 7,500,001 25

As an example of algorithm analysis let's look at two simple sorting algorithms

▪ Selection Sort and

▪ Insertion Sort

Calculate an approximate cost function for these two sorting algorithms

▪ By analyzing how many operations are performed by each algorithm

▪ This will include an analysis of how many times the algorithms' loops iterate

John Edgar 38

The array is divided into sorted part and unsorted parts Expand the sorted part by swapping the first unsorted element

with the smallest unsorted element

▪ Starting with the element with index 0, and

▪ Ending with the last but one element (index n – 1)

Requires two processes

▪ Finding the smallest element of a sub-array

▪ Swapping two elements of the array

John Edgar 39

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 1 3 5 7 21 27 29 23 19 13 31 15 17 9 25 11

The algorithm is on its fifth iteration Find the smallest element in arr[4:15]

smallest: 2119139

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 1 3 5 7 21 27 29 23 19 13 31 15 17 9 25 11

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 1 3 5 7 9 27 29 23 19 13 31 15 17 21 25 11 9

The array is divided into sorted part and unsorted parts Expand the sorted part by swapping the first unsorted element

with the smallest unsorted element

▪ Starting with the element with index 0, and

▪ Ending with the last but one element (index n – 1)

Requires two processes

▪ Finding the smallest element of a sub-array

▪ Swapping two elements of the array

John Edgar 40

The algorithm is on its fifth iteration Find the smallest element in arr[4:15]

smallest:

Swap smallest and first unsorted elements

void selectionSort(int arr[], int n){for(int i = 0; i < n-1; ++i){

int smallest = getSmallest(arr, i, n);swap(arr, i, smallest);

}}

John Edgar 41

int getSmallest(int arr[], int start, int end){ int smallest = start;for(int i = start + 1; i < end; ++i){

if(arr[i] < arr[smallest]){smallest = i;

}}return smallest;

}

void swap(int arr[], int i, int j){int temp = arr[i];arr[i] = arr[j];arr[j] = temp;

}

note: end is 1 past the last legal index



}}

John Edgar 42



}}return smallest;

}

void swap(int arr[], int i, int j){int temp = arr[i];arr[i] = arr[j];arr[j] = temp;

}

Swap always performs three operations

n-1 swaps 3(n-1)



}}

John Edgar 43



}}return smallest;

}

4(start-end-1)+4 operations

n-1 swaps 3(n-1)

called n-1 times

we can substitute n for end

the start index is the index of i in the calling function

the values of which are the sequence {0,1,2,…, n-2}

1:n-1, 2:n-1, …, n-1:n-1, average = (n-1+1) / 2 = n/2 average cost = 4(n/2)+4



}}

John Edgar 44

n-1 swaps 3(n-1)

called n-1 times (n-1)(4(n/2)+4)

Cost function: tselection sort = 2n2 - 2n + 10(n-1) + 2

for loop: 3(n-1)+2

= (n-1)(2n + 4)

= 2n2-2n+4(n-1)

The barometer operation for selection sort is in the loop that finds the smallest item

▪ Since operations in that loop are executed the greatest number of times

The loop contains four operations

▪ Compare i to end

▪ Compare arr[i] to smallest

▪ Change smallest

▪ Increment i

John Edgar 45

The barometer instructions

int getSmallest(arr[], start, end) smallest = startfor(i = start + 1; i < end; ++i)

if(arr[i] < arr[smallest])smallest = i

return smallest

Unsorted elements Barometer

n n-1

n-1 n-2

… …

3 2

2 1

1 0

n(n-1)/2

John Edgar 46

How is selection sort affected by the organization of the input?

▪ The only work that varies based on the input organization is whether or not smallest is assigned the value of arr[i]

What is the worst case organization? What is the best case organization? The difference between best case and worst case is

quite small

▪ (n-1)(3(n/2)) + 10(n-1) + 2 in the best case and

▪ (n-1)(4(n/2)) + 10(n-1) + 2 in the worst case

John Edgar 47

Ignoring leading constants, selection sort performs the following work

▪ n*(n – 1)/2 barometer operations, regardless of the original order of the input

▪ n – 1 swaps

The number of comparisons dominates the number of swaps

The organization of the input only affects the leading constant of the barometer operations

John Edgar 48

The array is divided into sorted part and unsorted parts The sorted part is expanded one element at a time

▪ By moving elements in the sorted part up one position until the correct position for the first unsorted element is found

▪ Note that the first unsorted element is stored so that it is not lost when it is written over by this process

▪ The first unsorted element is then copied to the insertion point

John Edgar 49

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 5 11 27 31 21 1 29 23 19 13 7 15 17 9 25 3

The algorithm is on its fourth iteration Find the correct position for arr[4]

temp: 21

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 5 11 27 31 21 1 29 23 19 13 7 15 17 9 25 3

The array is divided into sorted part and unsorted parts The sorted part is expanded one element at a time

▪ By moving elements in the sorted part up one position until the correct position for the first unsorted element is found

▪ Note that the first unsorted element’s value is stored so that it is not lost when it is written over by this process

▪ The first unsorted element is then copied to the insertion point

John Edgar 50

The algorithm is on its fourth iteration

Move up elements in the sorted part until the position for 21 is found

temp: 21

Find the correct position for arr[4]

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 5 11 27 31 31 1 29 23 19 13 7 15 17 9 25 3

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 5 11 27 27 31 1 29 23 19 13 7 15 17 9 25 3

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 5 11 21 27 31 1 29 23 19 13 7 15 17 9 25 3

How much work was performed by expanding the sorted part of the array by one element?

▪ The value 21 was stored in a variable, temp

▪ The values 27 and 31 were compared to 21▪ And moved up one position in the array

▪ The value 11 was compared to 21, but not moved

▪ The value of temp was written to arr[2]

How much work will be performed expanding the sorted part of the array to include the value 1?

How much work will be performed expanding the sorted part of the array to include the value 29?

index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

arr 5 11 21 27 31 1 29 23 19 13 7 15 17 9 25 3

John Edgar 51

void insertionSort(int arr[], int n){for(int i = 1; i < n; ++i){

temp = arr[i];int pos = i;// Shuffle up all sorted items > arr[i]while(pos > 0 && arr[pos - 1] > temp){

arr[pos] = arr[pos – 1];pos--;

} //while// Insert the current itemarr[pos] = temp;

}}

John Edgar 52

It depends on the values in the array

How often are they performed?

What are the barometer operations?

inner loop bodyhow many times?

void insertionSort(int arr[], int n){for(int i = 1; i < n; ++i){

temp = arr[i];int pos = i;// Shuffle up all sorted items > arr[i]while(pos > 0 && arr[pos - 1] > temp){

arr[pos] = arr[pos – 1];pos--;

} //while// Insert the current itemarr[pos] = temp;

}}

John Edgar 53

worst case: pos – 1 times for each iteration

outer loopn-1 times

What is the worst case organization? outer loop runs n-1 times: n * (n – 1) / 2

pos ranges from 1 to n-1; n/2 on average

Sorted

Elements

Worst-case Search

Worst-case Move

0 0 0

1 1 1

2 2 2

… … …

n-1 n-1 n-1

n(n-1)/2 n(n-1)/2

John Edgar 54

In the worst case the array is in reverse order Every item has to be moved all the way to the

front of the array

▪ The outer loop runs n-1 times

▪ In the first iteration, one comparison and move

▪ In the last iteration, n-1 comparisons and moves

▪ On average, n/2 comparisons and moves

▪ For a total of n * (n-1) / 2 comparisons and moves

John Edgar 55

The efficiency of insertion sort is affected by the state of the array to be sorted

What is the best case?

▪ In the best case the array is already completely sorted!

▪ No movement of any array element is required

▪ Requires n comparisons

John Edgar 56

What is the average case cost?

▪ Is it closer to the best case?

▪ Or the worst case?

If random data is sorted, insertion sort is usually closer to the worst case

▪ Around n * (n-1) / 4 comparisons

And what do we mean by average input for a sorting algorithm in anyway?

John Edgar 57

Quicksort is a more efficient sorting algorithm than either selection or insertion sort

▪ It sorts an array by repeatedly partitioning it

Partitioning is the process of dividing an array into sections (partitions), based on some criteria

▪ Big and small values

▪ Negative and positive numbers

▪ Names that begin with a-m, names that begin with n-z

▪ Darker and lighter pixels

John Edgar 59

John Edgar 60

Partition this array intosmall and big values using apartitioning algorithm

31 12 07 23 93 02 11 18

John Edgar 61


We will partition the arrayaround the last value (18),we'll call this value the pivot

31 12 07 23 93 02 11 18

Use two indices, one ateach end of the array, callthem low and high

18

John Edgar 62

31 12 07 23 93 02 11 18

arr[low] (31) is greater than the pivotand should be on the right, we need toswap it with something




John Edgar 63

31 12 07 23 93 02 11 18

arr[low] (31) is greater than the pivotand should be on the right, we need toswap it with something

arr[high] (11) is less than the pivot so swap with arr[low]




John Edgar 64

31 12 07 23 93 02 11 183111




John Edgar 65

12 07 23 93 02 18

increment low until it needs to be swapped with something

3111 12 07

then decrement high until it can be swapped with low




John Edgar 66

12 07 23 93 02 18

and then swap them

31230211 12 07

increment low until it needs to be swapped with something

then decrement high until it can be swapped with low




John Edgar 67

12 07 93 18

repeat this process until

31230211

high and low are the same




John Edgar 68

repeat this process until

high and low are the same

We'd like the pivot value to be in thecentre of the array, so we will swap itwith the first item greater than it

12 07 93 1831230211 9318




John Edgar 69

smalls bigspivot

12 07 9318 31230211




John Edgar 70

Use the same algorithm topartition this array into smalland big values

00 08 07 01 06 02 05 09

bigs!pivot

00 08 07 01 06 02 05 09

smalls

John Edgar 71

Or this one:

09 08 07 06 05 04 02 01

bigspivot

01 08 07 06 05 04 02 09

smalls

The quicksort algorithm works by repeatedly partitioning an array

Each time a sub-array is partitioned there is

▪ A sequence of small values,

▪ A sequence of big values, and

▪ A pivot value which is in the correct position

Partition the small values, and the big values

▪ Repeat the process until each sub-array being partitioned consists of just one element

John Edgar 72

The quicksort algorithm repeatedly partitions an array until it is sorted

▪ Until all partitions consist of at most one element

A simple iterative approach would halve each sub-array to get partitions

▪ But partitions are not necessarily the same size

▪ So the start and end indexes of each partition are not easily predictable

John Edgar 73

47 70 36 97 03 67 29 11 48 09 53

09 29 48 03 47 97

03 11 29 48 61

11

John Edgar 74

36 11 61 70

36 09 97

08 01 36

1109 03

0903

53

7047

29

11

53

48 61 977047 53

3629 48 61 977047 53

3629 48 61 977047 53

One way to implement quicksort might be to record the index of each new partition

But this is difficult and requires a reasonable amount of space

▪ The goal is to record the start and end index of each partition

▪ This can be achieved by making them the parameters of a recursive function

John Edgar 75

void quicksort(arr[], int low, int high){

if (low < high){

pivot = partition(arr, low, high);

quicksort(arr, low, pivot – 1);

quicksort(arr, pivot + 1, high);}

}

John Edgar 76

How long does Quicksort take to run?

▪ Let's consider the best and the worst case

▪ These differ because the partitioning algorithm may not always do a good job

Let's look at the best case first

▪ Each time a sub-array is partitioned the pivot is the exact midpoint of the slice (or as close as it can get)▪ So it is divided in half

▪ What is the running time?

John Edgar 77

John Edgar 78

08 01 02 07 03 06 04 05

bigspivot

04 01 02 03 05 06 08 07

smalls

First partition

John Edgar 79

big1pivot1

02 01 04 05 06 08

sm1

04 01 02 03 05 06 08 07

Second partition

0703

pivot1 pivot2

pivot2big2sm2

John Edgar 80

pivot1

02 03 04 05 06 07 08

Third partition

02 01 03 04 05 06 07 08

pivot1 donedonedone

01

John Edgar 81

Assume the best case – each partition splits its sub-array in half

qs(arr, 0, n-1)

sub-array size

n

qs(…) qs(…) qs(…) qs(…)

n/2

n/4

qs(…)qs(…)

# of recursive calls

1

2

3

…

log2(n) qs(…) qs(…) … 1

each level entails approximately n operations

there are approximately log2(n) levels

approximately log2(n) * n operations in total

Each sub-array is divided in half in each partition▪ Each time a series of sub-arrays are partitioned n

(approximately) comparisons are made

▪ The process ends once all the sub-arrays left to be partitioned are of size 1

How many times does n have to be divided in half before the result is 1?▪ log2(n) times

▪ Quicksort performs n * log2n operations in the best case

John Edgar 82

First partition

John Edgar 83

09 08 07 06 05 04 02 01

bigspivot

01 08 07 06 05 04 02 09

smalls

John Edgar 84

bigspivot

01 08 07 06 05 04 02 09

smalls

01 08 07 06 05 04 02 09

Second partition

John Edgar 85

bigspivot

01 02 07 06 05 04 08 09

01 08 07 06 05 04 02 09

Third partition

John Edgar 86

pivot

01 02 07 06 05 04 08 09

smalls

01 02 07 06 05 04 08 09

Fourth partition

John Edgar 87

bigspivot

01 02 04 06 05 07 08 09

01 02 07 06 05 04 08 09

Fifth partition

John Edgar 88

pivot

01 02 04 06 05 07 08 09

smalls

01 02 04 06 05 07 08 09

Sixth partition

John Edgar 89

pivot

01 02 04 05 06 07 08 09

01 02 04 06 05 07 08 09

Seventh partition!

John Edgar 90

Assume the worst case – each partition step results in a single sub-array

qs(arr, 0, n-1)

sub-array size

n

qs(…)

n-1

n-2

qs(…)

# of recursive calls

1

2

3

…

n qs(…) 1

each level entails on average n/2 operations

there are approximately n levels

approximately n2/2 operations in total

Every partition step ends with no values on one side of the pivot

▪ The array has to be partitioned n times, not log2(n) times

▪ So in the worst case Quicksort performs around n2

operations

The worst case usually occurs when the array is nearly sorted (in either direction)

John Edgar 91

With a large array we would have to be very, very unlucky to get the worst case

▪ Unless there was some reason for the array to already be partially sorted

The average case is much more like the best case than the worst case

There is an easy way to fix a partially sorted arrays to that it is ready for quicksort

▪ Randomize the positions of the array elements!

John Edgar 92

Calculation of a detailed cost function can be onerous and dependent on

▪ Exactly how the algorithm was implemented

▪ Implementing selection sort as a single function would have resulted in a different cost function

▪ The definition of a single discrete operation

▪ How many operations is this: mid = (low + high) / 2?

We are often interested in how algorithms behave as the problem size increases

John Edgar 94

There can be many ways to solve a problem▪ Different algorithms that produce the same result

▪ There are numerous sorting algorithms

Compare algorithms by their behaviour for large input sizes, i.e., as n gets large▪ On today’s hardware, most algorithms perform

quickly for small n Interested in growth rate as a function of n

▪ Sum an array: linear growth

▪ Sort with selection sort: quadratic growth

Measuring the performance of an algorithm can be simplified by

▪ Only considering the highest order term

▪ i.e. only consider the number of times that the barometer instruction is executed

▪ And ignoring leading constants

Consider the selection sort algorithm

▪ tselection sort = 2n2 - 2n + 10(n-1) + 2

▪ Its cost function approximates to n2

John Edgar 96

What are the approximate number of barometer

operations for the algorithms we looked at?

▪ Ignoring leading constants

Linear search: n

Binary search: log2n

Selection sort: n2

Insertion sort: n2

Quicksort: n(log2(n))

John Edgar 97

worst and average case


all cases


best and average case

What do we want to know when comparing two algorithms?

▪ Often, the most important thing is how quickly the time requirements increase with input size

▪ e.g. If we double the input size how much longer does an algorithm take?

Here are some graphs …

John Edgar 98

John Edgar 99

Arbitrary functions for the sake of illustration

Note that n is very small …

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0.00

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Note pairs of growth rates a constant distance apart

Exact counting of operations is often difficult (and tedious), even for simple algorithms

▪ And is often not much more useful than estimates due to the relative importance of other factors

O Notation is a mathematical language for evaluating the running-time of algorithms

▪ O-notation evaluates the growth rate of an algorithm

John Edgar 104

Cost Function: tA(n) = n2 + 20n + 100▪ Which term in the function is the most important?

It depends on the size of n▪ n = 2, tA(n) = 4 + 40 + 100

▪ The constant, 100, is the dominating term

▪ n = 10, tA(n) = 100 + 200 + 100▪ 20n is the dominating term

▪ n = 100, tA(n) = 10,000 + 2,000 + 100▪ n2 is the dominating term

▪ n = 1000, tA(n) = 1,000,000 + 20,000 + 100▪ n2 is still the dominating term

John Edgar 105

Big-O notation does not give a precise formulation of the cost function for a particular data size

It expresses the general behaviour of the algorithm as the data size n grows very large so ignores

▪ lower order terms and

▪ constants

A Big-O cost function is a simple function of n

▪ n, n2, log2(n), etc.

John Edgar 106

Express the number of operations in an algorithm as a function of n, the problem size

Briefly

▪ Take the dominant term

▪ Remove the leading constant

▪ Put O( … ) around it

For example, f(N) = 2n2 - 2n + 10(n-1) + 2

▪ i.e. O(n2)

Of course leading constants matter▪ Consider two algorithms

▪ f1(n) = 20n2

▪ f2(n) = 2n2

▪ Algorithm 2 runs ten times faster Let's consider machine speed

▪ If machine 1 is ten times faster than machine 2 it will run the same algorithm ten times faster

Big O notation ignores leading constants▪ It is a hardware independent analysis

Given a function T(n)

▪ Say that T(N) = O(f(n)) if T(n) is at most a constant times f(n)

▪ Except perhaps for some small values of n

Properties

▪ Constant factors don’t matter

▪ Low-order terms don’t matter

Rules

▪ For any k and any function g(n), k*g(n) = O(f(n))

▪ e.g., 5n = O(n)

An algorithm is said to be order f(n)

▪ Denoted as O(f(n))

The function f(n) is the algorithm's growth rate function

▪ If a problem of size n requires time proportional to n then the problem is O(n)

▪ e.g. If the input size is doubled so is the running time

John Edgar 110

An algorithm is order f(n) if there are positive constants k and m such that

▪ tA(n) k * f(n) for all n m▪ i.e. find constants k and m such that the cost function is less than or

equal to k * a simpler function for all n greater than or equal to m

If so we would say that tA(n) is O(f(n))

John Edgar 111

Finding a constant k | tA(n) k * f(n) shows that t is O(f(n))▪ e.g. If the cost function was n2 + 20n + 100 and we

believed this was O(n)▪ We claim to be able to find a constant k | tA(n) k * f(n)

for all values of n

▪ Which is not possible

For some small values of n lower order terms may dominate▪ The constant m addresses this issue

John Edgar 112

The idea is that a cost function can be approximated by another, simpler, function

▪ The simpler function has 1 variable, the data size n

▪ This function is selected such that it represents an upper bound on the value of tA(n)

Saying that the time efficiency of algorithm A tA(n) is O(f(n)) means that

▪ A cannot take more than O(f(n)) time to execute, and

▪ The cost function tA(n) grows at most as fast as f(n)

John Edgar 113

An algorithm’s cost function is 3n + 12

▪ If we can find constants m and k such that:

▪ k * n 3n + 12 for all n m then

▪ The algorithm is O(n)

Find values of k and m so that this is true

▪ k = 4, and

▪ m = 12 then

▪ 4n 3n + 12 for all n 12

John Edgar 114

A cost function is 2n2 - 2n + 10(n-1) + 2 If we can find constants m and k such that:

▪ k * n2 2n2 - 2n + 10(n-1) + 2 for all n m then

▪ The algorithm is O(n2)

Find values of k and m so that this is true

▪ k = 3, and

▪ m = 9 then

▪ 3n2 > 2n2 - 2n + 10(n-1) + 2 for all n 9

John Edgar 115

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After this point 3n2 is always going to be larger than 2n2 - 2n + 10(n-1) + 2

This is the graph of the selection sort cost function and 3n2

Demonstrating that selection sort is O(n2)

For all n 9, 3(n2) 2n2 - 2n + 10(n-1) + 2

i.e. k= 3 and m = 9

All these expressions are O(n)▪ n, 3n

▪ 61n + 5

▪ 22n – 5 All these expressions are O(n2)

▪ n2

▪ 9n2

▪ 18n2 + 4n – 53 All these expressions are O(n log n)

▪ n(log n)

▪ 5n(log 99n)

▪ 18 + (4n – 2)(log (5n + 3))

John Edgar 117

O(k * f) = O(f) if k is a constant

▪ e.g. O(23 * O(log n)), simplifies to O(log n)

O(f + g) = max[O(f), O(g)]

▪ O(n + n2), simplifies to O(n2)

O(f * g) = O(f) * O(g)

▪ O(m * n), equals O(m) * O(n)

▪ Unless there is some known relationship between m and nthat allows us to simplify it, e.g. m < n

John Edgar 118

Growth rate functions are typically one of the following

▪ O(1)

▪ O(log n)

▪ O(n)

▪ O(n*logn)

▪ O(n2)

▪ O(nk)

▪ O(2n)

John Edgar 119

We write O(1) to indicate something that takes a constant amount of time▪ Array look up

▪ Swapping two values in an array

▪ Finding the minimum element of an ordered array takes O(1) time▪ The minimum value is either the first or the last element of the array

▪ Binary or linear search best case Important: constants can be large

▪ So in practice O(1) is not necessarily efficient

▪ It tells us is that the algorithm will run at the same speed no matter the size of the input we give it

John Edgar 120

O(logn) algorithms run in logarithmic time

▪ Binary search average and worst case

▪ The logarithm is assumed to be base 2 unless specified

otherwise

Doubling the size of n doubles increases the number

of operations by one

Algorithms might be O(logn)

▪ If they are divide and conquer algorithms that halve the

search space for each loop iteration or recursive call

John Edgar 121

O(n) algorithms run in linear time

▪ Linear search

▪ Summing the contents of an array

▪ Traversing a linked list

▪ Insertion sort or bogo sort best case

Doubling the size of n doubles the number of operations Algorithms might be O(n)

▪ If they contain a single loop

▪ That iterates from 0 to n (or some variation)

▪ Make sure that loops only contain constant time operations

▪ And evaluate any function calls

John Edgar 122

O(nlogn)

▪ Mergesort in all cases

▪ Heap sort in all cases

▪ Quicksort in the average and best case

▪ O(nlogn) is the best case for comparison sorts

▪ We will not prove this in CMPT 225

Growth rate is faster than linear but still slow compared to O(n2)

Algorithms are O(nlogn)

▪ If they have one process that is linear that is repeated O(logn) times

John Edgar 123

O(n2) algorithms run in quadratic time

▪ Selection sort in all cases

▪ Insertion sort in the average and worst case

▪ Bubble sort in all cases

▪ Quicksort worst case

Doubling the size of n quadruples the number of operations

Algorithms might be O(n2)

▪ If they contain nested loops

▪ As usual make sure to check the number of iterations in such loops

▪ And that the loops do not contain non-constant function calls

John Edgar 124

O(nk) algorithms are referred to as running in

polynomial time

▪ If k is large they can be very slow

Doubling the size of n increases the number of

operations by 2k

John Edgar 125

O(2n) or O(kn) algorithms are referred to as running in exponential time

Very slow, and if there is no better algorithm implies that the problem is intractable

▪ That is, problems of any reasonable size cannot be solved in a reasonable amount of time▪ Such as over the lifetime of the human race …

O(n!) algorithms are even slower

▪ Traveling Salesman Problems (and many others)

▪ Bogo sort in the average case

John Edgar 126

The O notation growth rate of some algorithms varies depending on the input

Typically we consider three cases:

▪ Worst case, usually (relatively) easy to calculate and therefore commonly used

▪ Average case, often difficult to calculate

▪ Best case, usually easy to calculate but less important than the other cases▪ Relying on the input having the best case organization is not

generally a good idea

John Edgar 127

(omega) notation

▪ Gives a lower bound

▪ Whereas O notation is an upper bound

▪ There exists some constant k, such that k * f(n) is a lower

bound on the cost function g(n)

(theta) notation

▪ Gives an upper and lower bound

▪ k1 * f(n) is an upper bound on g(n) and k2 * f(n) is a lower

bound on g(n)

John Edgar 128

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O(n2)

(n2)

CMPT 225 - Simon Fraser University · 2017. 10. 15. · Choosing an appropriate algorithm can make...

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